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The problem is to find the area of the region enclosed by the given parabola and line:

$y=-x^2+3x-1$

$y=x-2$

I tried solving the definite integral, but the roots (points of intersection of the parabola and line) on the graph were too weird and I ended up with the wrong solution. Does anyone know a more efficient method?

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This is a method where you do not need to find the actual roots from the graph, from which the problem of "weird points of intersection" will be completely eradicated. Trust me.

From $-x^2+3x-1=x-2$, $x^2-2x-1=0$.

Letting $a$ and $b$ (where $a<b$) be the solutions of $x^2-2x-1=0$, $a+b=-2$ and $ab=1$, due to the root-coefficient relationship.

Therefore, the area, $S = \int_a^b\left[-x^2+3x-1-\left(x-2\right)\right]dx = -\int_a^b\left(x^2-2x-1\right)dx = -\int_a^b\left(x-a\right)\left(x-b\right)dx = \frac{1}{6}\left(b-a\right)^3$

From $(b-a)^2=(a+b)^2-4ab=8$, $b-a=2\sqrt{2}$

Therefore, $\left(b-a\right)^3=16\sqrt{2}$.

Thus, the area, $S=\frac{8\sqrt{2}}{3}$.

And this answer was able to be posted quickly due to my MathJax friend, Desmos.

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  • $\begingroup$ Thank you. How did you evaluate that to 1/6(b-a)^3? Did you just solve it? $\endgroup$ – hello world Jan 28 '18 at 16:32
  • $\begingroup$ $-\int_a^b\left(x-a\right)\left(x-b\right)dx = \frac{1}{6}\left(b-a\right)^3$ You can just solve it, However $\int_a^b\left(x-a\right)\left(x-b\right)dx = -\frac{1}{6}\left(b-a\right)^3$ $\endgroup$ – Yash Jain Jan 28 '18 at 16:35
  • $\begingroup$ I will upvote now because I have enough reputation, and your answer was useful. $\endgroup$ – hello world Jan 28 '18 at 23:38
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This problem has a nice solution without Calculus.

enter image description here

Let $P$ and $Q$ the intersections between the given parabola and the given line. Let $T$ be the intersection of the tangents at $P$ and $Q$. By Archimedes' lemma, the area of the wanted parabolic segment is just two thirds of the area of $PQT$. We have $T=(1;3)$ and $PQ=4$ by straightforward computations. The distance of $T$ from the $PQ$ line equals $2\sqrt{2}$, hence $[PQT]=\frac{1}{2}\cdot 4\cdot 2\sqrt{2}=4\sqrt{2}$ and the wanted area equals $\color{blue}{\frac{8}{3}\sqrt{2}}$.

Yet another interesting approach comes from noticing that Simpson's rule is an exact quadrature rule for quadratic polynomials, hence the given problem can be solved by just computing the coordinates of $P,Q$ and by evaluating $-x^2+3x-1$ at $x=1$, since that is the average of the abscissas of $P$ and $Q$.

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  • $\begingroup$ I will upvote now because I have enough reputation, and your answer was useful. $\endgroup$ – hello world Jan 28 '18 at 23:38
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at the first step you must compute the intersection Points of both functions this leads to the equation $$-x^2+2x+1=0$$ can you solve this? for your Control we get $$x_1=1-\sqrt{2}$$ $$x_2=1+\sqrt{2}$$ and then the searched area is given by $$A=\int_{x_1}^{x_2} -x^2+3x-1-(x-2)dx$$

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  • $\begingroup$ and you should post what you have tried $\endgroup$ – Dr. Sonnhard Graubner Jan 28 '18 at 16:22
  • $\begingroup$ it is not a comment, i want do know if the results are ok $\endgroup$ – Dr. Sonnhard Graubner Jan 28 '18 at 16:25
  • $\begingroup$ I have already done this method, I just want to find a more efficient method in which I do not have to calculate these values 1-sqrt(2) and 1+sqrt(2) $\endgroup$ – hello world Jan 28 '18 at 16:26
  • $\begingroup$ but you will Need These Terms! $\endgroup$ – Dr. Sonnhard Graubner Jan 28 '18 at 16:27
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    $\begingroup$ yeah ok, i know this method! but the next Problem can be complicated and he/she will Need that method $\endgroup$ – Dr. Sonnhard Graubner Jan 28 '18 at 16:29
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This can be done using your suggested integration method. To find the intersections we set $x^{2}-3x+1=2-x$ so we have $x^{2}-2x-1=0$ so our limits of integration are the solutions to this quadratic namely $x=1\pm\sqrt{2}$. Now to find the area we want the integral $\int_{1-\sqrt{2}}^{1+\sqrt{2}}-x^{2}+2x+1dx=[\frac{x^{3}}{3}+x^{2}+x]_{1-\sqrt{2}}^{1+\sqrt{2}}=\frac{8\sqrt{2}}{3}$

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  • $\begingroup$ I have already tried this method. $\endgroup$ – hello world Jan 28 '18 at 16:33
  • $\begingroup$ If you simplify the integration (after substituting the values in) you will get to $\frac{8\sqrt{2}}{3}$ the same answer as the alternative method. $\endgroup$ – pilgrim Jan 28 '18 at 16:38
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This is a simple integration problem where the required steps are:

  1. define your domain by:

    a. finding the graphs intersection points.

    b. understanding which graph is above the other (simple plot always helps).

  2. double-integrate "1" over your domain.

In your case, you will have $1-\sqrt{2}$ and $1+\sqrt{2}$ as your intersection points, and within this range the parabola lies above the line, so you have:

$$\int_{1-\sqrt{2}}^{1+\sqrt{2}}\int_{x-2}^{-x^2+3x-1} 1 dy dx$$ = $$ \int_{1-\sqrt{2}}^{1+\sqrt{2}} (-x^2+3x-1)-(x-2) dx$$ =

$$ (-x^3 / 3 + x^2 + x)\Bigr|_{1-\sqrt{2}}^{1+\sqrt{2}} $$ $\approx$ 3.77

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